Express 6x^2+12x+7 in completed square form. Hence, show that f(x) = 2x^3+6x^2+7x-3 is a strictly increasing function for all values of x.
To complete the square in the quadratic, first pull out a common factor from the first two terms to make sure the x2 term has a coefficient of 1. 6x2+12x+7 = 6(x2+2x)+7. Now express x2+2x as a square take away a constant. To do this, we halve 2 and notice x2+2x=(x+1)2-1. So we have 6x2+12x+7 = 6[(x+1)2-1]+7, and multiplying by 6 and cleaning up we get 6x2+12x+7 = 6(x+1)2+1. To find out if f is increasing, we need to differentiate it: f'(x)=6x2+12x+7. Using the above part, f'(x)=6(x+1)2+1. Squares are always non-negative so 6(x+1)2 >= 0, and so f'(x) > 0. As f'(x) is strictly positive, f(x) is a strictly increasing function.
